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Calc 3 10.7

 

<== Calculus 3
Number Question Answer
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1. Find the domain of the vector functions, \mathbf r(t), listed below.
a) \mathbf r(t) = \langle \ln(7 t), \sqrt{t  + 9}, rac{1}{\sqrt{20 - t}}
angle

b) \mathbf r(t) = \langle \sqrt{t  - 7 }, \sin(1 t), t^2 
angle

c) \mathbf r(t) = \langle \ e^{-7 t}, rac{t}{\sqrt{t^2 - 1}}, t^{1/3} 
angle

A) (0, 20)

B) [7, &infin]

C) (-&infin, -1) U (1, &infin)

The work
2. Find the domain of the vector function

\mathbf r(t) = \langle \ln(6 t), \sqrt{t  + 17}, rac{1}{\sqrt{13 - t}}
angle

(0, 13)
The work
3. 	extrm{Let }\displaystyle \mathbf{r}(t)= (\sqrt{t+2})\,\mathbf{i} + 
                     \left(rac{t^2-{1}}{t-1}
ight)\,\mathbf{j} +
                      \sin(2\pi t)\,\mathbf{k}.
Then
\displaystyle \lim_{ t 
ightarrow 1} \mathbf{r}(t) i + j + k.
\displaystyle \lim_{ t 
ightarrow 1} \mathbf{r}(t) 1.7321 i + 2j + 0 k.
The work
4. Find the limit:

\lim_{t 	o 0}\langle
    rac{e^{-2 t} - 1}{t}, rac{t^{3}}{t^{4} - t^{3}},rac{-8}{4 + t}

angle

&lang, , &rang
&lang -2, -1, -2 &rang
The work
5. Find a vector function that represents the curve of intersection of the paraboloid z = 8 x^2 + 3 y^2 and the cylinder y = 4 x^2. Use the variable t for the parameter.
\mathbf r(t) = \langle t, = &lang t, , &rang
\mathbf r(t) = \langle t, = &lang t, , &rang
The work
6. Find the derivative of the vector function

r(t) = \ln(11 - t^2) \mathbf{i} + 
    \sqrt{13 + t}\mathbf{j}  - 3 e^{5 t}\mathbf{k}
\mathbf r'(t) = \langle = &lang , , &rang
\mathbf r'(t) = \langle = &lang , , &rang
The work
7. For the given position vectors \mathbf{r}(t),
compute the (tangent) velocity vector \mathbf{r}'(t) for the given value of t .

A) r(t) = (cos(2t), sin(2t)).
Then \mathbf{r}'(rac{\pi}{4})= ( , )?

B) \displaystyle 	extrm{Let }  {\bf{r}}(t)= (t^2,t^3).
Then r'(4) = ( , )?

C) \displaystyle 	extrm{Let }  \mathbf{r}(t)= e^{2t}\mathbf{i}+ e^{-4t}\mathbf{j}+
  t\mathbf{k}.
Then r'(4) = i + j + k ?

A) \mathbf{r}'(rac{\pi}{4})= ( -2, 0)

B) r'(4) = ( 8, 48)

C) r'(4) =( 5961.92, -4.5*10^-7, 1)

The work
8. Find the unit tangent vector at the indicated point of the vector function

\mathbf r(t) = e^{10 t}\cos t \mathbf{i} + 
    e^{10 t}\sin t \mathbf{j} + e^{10 t}\mathbf{k}
\mathbf T(\pi/2) = \langle &lang , , &rang
\mathbf T(\pi/2) = \langle &lang -0.0705, 0.705, 0.705 &rang
The work
9. For the given position vectors r (t) compute the unit tangent vector T (t) for the given value of t .

A) r (t) = (cos(5t), sin(5t)).
Then \mathbf{T}(rac{\pi}{4}) ( , )

B) 	extrm{Let } \mathbf{r}(t)= (t^2,t^3).
Then T (3)= ( , )

C) 	extrm{Let } \mathbf{r}(t)= e^{5t}\mathbf{i}+ e^{-3t}\mathbf{j}+
  t\mathbf{k}.
Then T (-5) = i + j + k .

A) ( 0.707, -0.707)

B) ( 0.2169, 0.9762)

C) 0i + -1j + 1.02E-07k

The work
10. Find the derivative of the vector function
\mathbf r(t) = t\mathbf a 	imes (\mathbf b + t\mathbf c), where
a = &lang 1, 1, 5 &rang, b = &lang -1, 3, -4 &rang, and c = &lang -5, -2, -1 &rang
r'(t) = &lang , , &rang
r'(t) = &lang -19 + 18t, -1 - 48t, 4 + 6t &rang
The work
11. Consider the vector function
\mathbf r(t) = \langle t, t^{2}, t^{2}
angle Compute
r'(t) = &lang , , &rang
T(1) = &lang , , &rang
r''(t) = &lang , , &rang
\mathbf r'(t)	imes \mathbf r''(t) = \langle &lang , , &rang
r'(t) = &lang 1, 2t, 2t&rang
T(1) = &lang 1/3, 2/3, 2/3 &rang
r''(t) = &lang 0, 2, 2&rang
\mathbf r'(t)	imes \mathbf r''(t) = \langle &lang 0, -2, 2&rang
The work
12. Find parametric equations for the tangent line at the point
(\cos(rac{2\pi}{6}) ,\sin(rac{2\pi}{6}) ,rac{2\pi}{6} ) on the curve x = cos(t), y = sin(t), z = t
x(t) =
y(t) =
z(t) =
x(t) =
y(t) =
z(t) =
The work
13. Find the parametric equations for the tangent line to the curve

x =
    t^{3} - 1, y = t^{3} + 1, z = t^{1}
at the point (0, 2, 1). Use the variable t for your parameter.
x = ,
y = ,
z =
x = 3t,
y = 2 + 3t,
z = 1 + t
The work
14. Evalute
\int_{0}^{7}(t\mathbf{i}+ t^2\mathbf{j}+t^3\mathbf{k})dt i + j + k.
24.5 i + 114.33 j + 600.25 k
The work
15. If r(t) = cos(-1t)i + sin( -1t) j + 0t k
compute r'(t)= i + j + k
and \int{\bf{r}}(t)\, dt= i + j + k + C
with C a constant vector.
r'(t)= sin(-t) i - cos(-t) j + 0 k
\int{\bf{r}}(t)\, dt= -sin(-t) i + cos(-t)j + 0 k + C
The work
16. Let \mathbf{c}_1(t) = (e^{-2t}, \sin(3t), -3t^3), and \mathbf{c}_2(t) = (e^{-t}, \cos(3t), -5t^3)

\displaystyle rac{d}{dt}\left[ \mathbf{c}_1(t) \cdot  \mathbf{c}_2(t)
ight]  =
\displaystyle rac{d}{dt}\left[ \mathbf{c}_1(t) 	imes  \mathbf{c}_2(t)
ight]  = i + j + k

\displaystyle rac{d}{dt}\left[ \mathbf{c}_1(t) \cdot  \mathbf{c}_2(t)
ight]  =
\displaystyle rac{d}{dt}\left[ \mathbf{c}_1(t) 	imes  \mathbf{c}_2(t)
ight]  = i + j + k
The work
17. Find the tangent vector to the curve c(t) = ( g(t), h(t), f(g(t), h(t)) ) on the surface z = f(x,y) = x^2 + \sin(xy) when t = 1, and where x = 3t + 7 and y = 6 t^2 + 4 t.

( , , ).

( 3, 16, 223.841 )
The work